Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be positive reals such that $$a_1 + b_1 = a_2 + b_2 = \ldots + a_n + b_n$$ and $$\sqrt[n]{\dfrac{a_1a_2\ldots a_n}{b_1b_2\ldots b_n}} \ge n.$$ Prove that $$\sqrt[n]{\dfrac{a_1a_2\ldots a_n}{b_1b_2\ldots b_n}} \ge \dfrac{a_1+a_2+\ldots+a_n}{b_1+b_2+\ldots+b_n}.$$